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From: Aatu Koskensilta on 31 Mar 2010 06:59
zuhair <zaljohar(a)gmail.com> writes:

> If we have a set theory T that disprove choice, and we have two
> sub-theories T1, T2 of T, each consistent with choice (do not
> disprove, neither prove choice) is there a possibility that the theory
> "T1+T2" disprove choice?

Sure, take a theorem A of T which is not a logical truth and does not
logically imply the negation of choice, and have T1 consist just of "if
A, choice fails" and T2 of A.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: zuhair on 31 Mar 2010 07:14
On Mar 30, 4:39 pm, zuhair <zaljo...(a)gmail.com> wrote:
> If we have a set theory T that disprove choice, and we
> have two sub-theories T1, T2  of T, each consistent with choice (do
> not disprove, neither prove choice) is there a possibility that
> the theory "T1+T2" disprove choice?
>
> Zuhair

Well according to Mr.R.Holmes

NFP + NFU is NF

while neither NFP nor NFU disprove choice, NF does.

Zuhair
From: Tim Little on 31 Mar 2010 21:06
On 2010-03-30, zuhair <zaljohar(a)gmail.com> wrote:
> If we have a set theory T that disprove choice, and we
> have two sub-theories T1, T2 of T, each consistent with choice (do
> not disprove, neither prove choice) is there a possibility that
> the theory "T1+T2" disprove choice?

Yes, for example take T1 = ZF and T2 = an axiom of negation of well
ordering. T2 alone proves almost nothing; certainly not the negation
of choice.

A simpler example, though trivial: T1 consists of a single axiom P, T2
consists only of ~P. Neither T1 nor T2 disproves choice, but as T is
necessarily inconsistent, it does disprove choice (and everything
else).


- Tim
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